which is larger: $\sin(4^\circ)$ or $2\sin(2^\circ)$?

hint: What's the length of the red line?

Here is a simple way to see this. The chord function is defined as $$\text{crd}(x) := 2\sin(x/2).$$ In order to show that $$\sin(2x)<2\sin(x)$$ for all $\,x\,$ it is equivalent to showing that $$\text{crd}(2x)<2\,\text{crd}(x)$$ for all $\,x\,$ but this is just the triangle inequality applied to an isoceles triangle inscribed in a circle. The base of the triangle is the chord of twice the angle of the other two equal sides which are chords the the same angle.

The sine wave is concave down between $0$ and $180^\circ$, so the line that passes through the origin and $(2^\circ,\sin 2^\circ)$ will pass above $(4^\circ,\sin4^\circ)$. Thus, $\sin 4^\circ<2\sin 2^\circ$.